Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4 minutes to raise the temperature of 0.5 kg water by 40K. This heater is replaced by a new heater having two wires of the same material, each of length L and diameter 2d. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40 K?

Options

A. 4 if wires are in parallel
B. 2 if wires are in series
C. 1 if wires are in series
D. 0.5 if wires are in parallel

Answer

B, D

Explanation

The length and material are same in both cases. The resistance of coil will be inversely proportional to area of cross section. In the second case, the diameter is double, which means the area becomes four times. Hence, the resistance of each wire in second case is of resistance in the first case. Let R be the resistance of coil in the first case then the amount of heat generated in the first case can be written as follows:
`Q= (V^(2))/(R )t` ...(i)
In the second case, there are two wires of resistance R/4 each. When the two are connected in series then equivalent resistance becomes R/2 and let it take time `t_(S)` to generate the same amount of heat. Then we can write as follows:
`Q= (V^(2))/(R//2) t_(S) rArr Q = (2V^(2))/(R ) t_(S)` ...(ii)
When the two resistances are connected in parallel then effective resistance becomes R/8. Let it takes time `t_(P)` to generate the same amount of heat then we can write the following equation.
`Q= (V^(2))/(R//8) t_(P) rArr Q = (8V^(2))/(R ) t_(P)`...(iii)
From above equations we can see that:
`t_(S) = t//2 = 4//2 =2` minutes
`t_(P) = t//8= 4//8 = 0.5` minutes
Hence, when the coils are connected in series then they will take two minutes and when connected in parallel then they will take 0.5 minutes to heat the water. Hence, option (b) and (d) are correct.